If 3 + 4i is a root of the equation {x^2} + p | vedclass

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(D) Since the coefficients $p$ and $q$ of the quadratic equation ${x^2} + px + q = 0$ are real,complex roots must occur in conjugate pairs.Given that

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$p = -6, q = 25$

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(D) Since the coefficients $p$ and $q$ of the quadratic equation ${x^2} + px + q = 0$ are real,complex roots must occur in conjugate pairs.Given that $3 + 4i$ is a root,its conjugate $3 - 4i$ must also be a root of the equation.For a quadratic equation ${x^2} + px + q = 0$,the sum of the roots is given by $-p$ and the product of the roots is given by $q$.Sum of the roots $= (3 + 4i) + (3 - 4i) = 6$.Therefore,$-p = 6$,which implies $p = -6$.Product of the roots $= (3 + 4i)(3 - 4i) = {3^2} - {(4i)^2} = 9 - 16({i^2}) = 9 - 16(-1) = 9 + 16 = 25$.Therefore,$q = 25$.Thus,the values are $p = -6$ and $q = 25$.

If $3 + 4i$ is a root of the equation ${x^2} + px + q = 0$ ($p, q$ are real numbers),then

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